Lesson Notes By Weeks and Term v3 - Senior Secondary 2
Permutation & combinations
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Permutation & combinations
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Subject: Further Mathematics
Class: Senior Secondary 2
Term: 1st Term
Week: 4
Theme: Statistics
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Under stand concepts of permutation and combination Solve problems in cyclic permutation Solve other problems on permutation Solve various types of problems on combination
2. 1. Fundamental Principle of Counting (Multiplication Rule) This principle states that if an event can occur in 'm' ways, and after it has occurred, another event can occur in 'n' ways, then the two events can occur in m x n ways. This extends to any number of events.
Example 1: A student has 3 different traditional attires (Agbada, Senator, Ankara), 2 different caps (Fila, Kufi), and 2 different types of sandals. The number of ways the student can dress up is 3 x 2 x 2 = 12 ways. 2.
2. Factorial Notation (n!) The factorial of a non-negative integer 'n', denoted by n!, is the product of all positive integers less than or equal to n.
Definition: n! = n x (n-1) x (n-2) x ... x 3 x 2 x 1 Special Case: 0! = 1 (by definition, important for formulas).
Example 2: 3! = 3 x 2 x 1 = 6 5! = 5 x 4 x 3 x 2 x 1 = 120 Calculate 7! / 4!: 7! / 4! = (7 x 6 x 5 x 4 x 3 x 2 x 1) / (4 x 3 x 2 x 1) = 7 x 6 x 5 =
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0. Alternatively, 7! / 4! = 7 x 6 x 5 x 4! / 4! = 7 x 6 x 5 = 210. 2.
3. Permutation Permutation refers to the different arrangements of a set of objects where the order of arrangement matters. It is about ordering elements. 2.3.
1. Linear Permutation of Distinct Objects Arranging all 'n' distinct objects: The number of ways to arrange 'n' distinct objects in a line is n!.
Example 3: How many ways can the letters of the word 'LAGOS' be arranged? There are 5 distinct letters. So, the number of arrangements = 5! = 5 x 4 x 3 x 2 x 1 = 120 ways. Arranging 'r' objects from 'n' distinct objects: The number of permutations of 'r' objects chosen from 'n' distinct objects is denoted by P(n, r) or nPr.
Formula: P(n, r) = n! / (n-r)!
Example 4: In how many ways can 3 prizes be awarded to 10 students if each student can receive at most one prize? This is a permutation because the order of receiving prizes (1st, 2nd, 3rd) matters. n = 10 (total students), r = 3 (prizes to be awarded). P(10, 3) = 10! / (10-3)! = 10! / 7! = 10 x 9 x 8 = 720 ways. 2.3.
2. Permutation with Repetition (Identical Objects) When some of the 'n' objects are identical, the formula for permutation is adjusted to avoid counting identical arrangements multiple times.
Formula: If there are 'n' objects where p objects are of one kind, q objects of a second kind, r objects of a third kind, and so on, then the number of distinct permutations is n! / (p!q!r!...).
Example 5: How many distinct ways can the letters of the word 'BENUE' be arranged? There are 5 letters in total (n=5). The letter 'E' appears 2 times (p=2). All other letters (B, N, U) appear once. Number of distinct arrangements = 5! / 2! = (5 x 4 x 3 x 2 x 1) / (2 x 1) = 120 / 2 = 60 ways.
Example 6: How many distinct arrangements of the letters of the word 'ABUJA' are possible? n=
5. The letter 'A' appears 2 times (p=2). Number of arrangements = 5! / 2! = 60 ways. 2.3.
3. Cyclic Permutation Cyclic permutation refers to arrangements of objects in a circle. In a circular arrangement, there is no distinct "start" or "end" position. Arranging 'n' distinct objects in a circle: Formula: (n-1)!
Explanation: If we fix one object's position, the remaining (n-1) objects can be arranged in (n-1)! ways relative to the fixed object.
Example 7: How many ways can 6 people be seated around a round dining table? n =
6. Number of ways = (6-1)! = 5! = 5 x 4 x 3 x 2 x 1 Cyclic permutation refers to arrangements of objects in a circle. In a circular arrangement, there is no distinct "start" or "end" position. Arranging 'n' distinct objects in a circle: Formula: (n-1)!
Explanation: If we fix one object's position, the remaining (n-1) objects can be arranged in (n-1)! ways relative to the fixed object.
Example 7: How many ways can 6 people be seated around a round dining table? n =
6. Number of ways = (6-1)! = 5! = 5 x 4 x 3 x 2 x 1 = 120 ways. Arranging 'n' distinct objects in a circle where clockwise and anti-clockwise arrangements are considered the same: Formula: (n-1)! / 2 Explanation: This typically applies to arrangements where the object itself has no orientation (e.g., beads on a necklace, flowers in a garland). If turning the arrangement over results in the same sequence, we divide by
2. Example 8: How many ways can 7 different coloured beads be arranged to form a necklace? n =
7. Since turning the necklace over results in the same arrangement, we divide by
2. Number of ways = (7-1)! / 2 = 6! / 2 = (720) / 2 = 360 ways. 2.
4. Combination Combination refers to the different selections of a set of objects where the order of selection does not matter. It is about choosing elements. Selecting 'r' objects from 'n' distinct objects: The number of combinations of 'r' objects chosen from 'n' distinct objects is denoted by C(n, r) or nCr or $\binom{n}{r}$.
Formula: C(n, r) = n! / (r! (n-r)!)
Relationship to Permutation: C(n, r) = P(n, r) / r!
Properties: C(n, r) = C(n, n-r) C(n, 0) = 1 (There's one way to choose 0 items: choose none.) C(n, n) = 1 (There's one way to choose all 'n' items.)
Example 9: A local government is forming a committee of 4 members from a group of 10 qualified candidates. How many different committees can be formed? Here, the order in which members are selected does not matter (a committee of A, B, C, D is the same as B, A, D, C). So, it is a combination. n = 10 (total candidates), r = 4 (members to be selected). C(10, 4) = 10! / (4! (10-4)!) = 10! / (4! 6!) = (10 x 9 x 8 x 7 x 6!) / ((4 x 3 x 2 x 1) x 6!) = (10 x 9 x 8 x 7) / (4 x 3 x 2 x 1) = (10 x 3 x 7) = 210 different committees.
Example 10: A basket contains 5 oranges, 4 mangoes, and 3 apples. In how many ways can a person select 2 oranges, 2 mangoes, and 1 apple? This involves selecting items from different categories, so we apply the multiplication rule along with combination for each category.
Ways to select 2 oranges from 5: C(5, 2) = 5! / (2! 3!) = (5 x 4) / (2 x 1) = 10 Ways to select 2 mangoes from 4: C(4, 2) = 4! / (2! 2!) = (4 x 3) / (2 x 1) = 6 Ways to select 1 apple from 3: C(3, 1) = 3! / (1! 2!) = 3 / 1 = 3 Total ways = C(5, 2) x C(4, 2) x C(3, 1) = 10 x 6 x 3 = 180 ways. 3.
1. Teacher Activities: Introduction (5-10 minutes): Begin by posing real-life scenarios that differentiate between order matters and order doesn't matter (e.g., "Selecting a class prefect and a vice prefect" vs. "Selecting two students for a committee"). Introduce the terms "permutation" and "combination" as the mathematical tools to solve such counting problems. Factorial Notation Explanation (10 minutes): Explain the concept of factorial notation (n!) with basic examples (e.g., 3!, 5!). Emphasize 0! = 1 and its importance. Practice simplification problems involving factorials (e.g., 7!/4!).
Permutation Concepts (20-25 minutes): Define permutation and explain its formula P(n, r) = n! / (n-r)!. Work through examples of linear permutation for distinct objects (e.g., arranging letters of a word, assigning positions). Introduce permutation with repetition (for identical objects) using the formula n! / (p!q!r!...) with examples like arranging letters of 'BENUE' or 'COMMISSION'. Explain cyclic permutation, providing examples of seating around a round table (n-1)! and arrangements where clockwise/anti-clockwise are identical (e.g., beads on a necklace, (n-1)!/2).
Combination Concepts (20-25 minutes): Define combination and clearly differentiate it from permutation using practical examples. Introduce the formula C(n, r) = n! / (r! (n-r)!). Work through examples of combination problems (e.g., selecting committees, choosing lottery numbers, selecting items from different groups). Highlight the properties of combinations, such as C(n, r) = C(n, n-r). Guided Practice and Problem Solving (20 minutes): Provide scaffolded problems covering all aspects (factorial, linear P, cyclic P, C) for students to solve. Circulate around the classroom, providing support, clarification, and correcting misconceptions. Invite students to present their solutions on the board.
Wrap-up and Summary (5 minutes): Summarize the key differences between permutation and combination. Reiterate the formulas and their applications. Assign independent practice questions. 3.
2. Student Activities: Actively participate in class discussions and question-and-answer sessions. Work individually or in small groups to solve problems presented by the teacher. Take notes on definitions, formulas, and worked examples. Engage in peer-to-peer learning by discussing problem-solving strategies. Present solutions to problems on the board, explaining their reasoning. Ask clarifying questions when concepts are unclear. Complete assigned practice problems and homework. 3 x 2 x 1) = (15 x 14 x 13 x 12) / (4 x 3 x 2 x 1) = 15 x 7 x 13 / 1 = 1365 ways. b) 3 specific players must be included. This means the coach needs to select (11 - 3) = 8 more players. The remaining players available for selection are (15 - 3) = 12 players. So, we need to choose 8 players from the remaining
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2. C(12, 8) = 12! / (8! (12-8)!) = 12! / (8! 4!) = (12 x 11 x 10 x 9 x 8!) / (8! x 4 x 3 x 2 x 1) = (12 x 11 x 10 x 9) / (4 x 3 x 2 x 1) = 11 x 5 x 9 = 495 ways.
Commentary: This combination problem includes a common constraint where certain items must be included, simplifying the selection pool.
Logistics and Scheduling in Transportation (e.g., Commercial Bus Routes): Application: Transport companies in cities like Lagos or Abuja need to plan optimal routes and schedules for their commercial buses (e.g., BRT). If there are multiple bus stops and a specific number of buses, permutations can help determine the various sequences in which buses can visit stops, especially if the order of stops affects efficiency or cost. Combinations might be used to select a subset of routes to operate on a particular day.
Integration: Students can be given a scenario involving a hypothetical transport company with a set number of bus stops and buses, and asked to calculate possible routes or crew assignments. Committee Formation and Resource Allocation in Local Governance: Application: In Nigerian local government areas, various committees (e.g., education, health, youth development) are formed from a pool of community leaders or civil servants. Combinations are directly applied to calculate the number of ways to select members for these committees, ensuring fair representation or specific expertise. If positions (e.g., Chairman, Secretary) are assigned, then permutations come into play.
Integration: Students can discuss a scenario where a local government council needs to form a task force for a community project, and apply combinations to determine selection possibilities given specific criteria (e.g., a certain number of men and women, or professionals from different fields). Security and Digital Passwords (e.g., ATM PINs, Mobile Banking Apps): Application: Permutations are foundational to understanding the strength and number of possible combinations for security features like ATM PINs (4 distinct digits), phone unlock patterns, or online banking passwords. A higher number of permutations means greater security.
Integration: Teachers can engage students in a discussion about why using simple, repetitive PINs is risky, by demonstrating how permutations calculate the vast number of possibilities available even with just 4 digits when order matters and repetition is allowed/disallowed. Students can calculate the possible unique combinations for different length PINs or character sets.